Thread: Slope of the tangent line

The slope of the tangent live to a curve is given by f`(x)=3x^2+5 if the point
(1,2) is on the curve, find the equation of the curve.
if i integrate this i get x^3+5x+C however the result has a constant of -4, that is => x^3+5x-4
So how would i go about finding the constant ?

Originally Posted by xterminal01

The slope of the tangent live to a curve is given by f`(x)=3x^2+5 if the point
(1,2) is on the curve, find the equation of the curve.
if i integrate this i get x^3+5x+C however the result has a constant of -4, that is => x^3+5x-4
So how would i go about finding the constant ?

plug in the initial condition ...

Originally Posted by xterminal01

The slope of the tangent live to a curve is given by f`(x)=3x^2+5 if the point
(1,2) is on the curve, find the equation of the curve.
if i integrate this i get x^3+5x+C however the result has a constant of -4, that is => x^3+5x-4
So how would i go about finding the constant ?

(1,2) <---this is your initial condition